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The Baire category theorem implies that a nonempty complete metric space without isolated points must be uncountable. In many situations I have encountered, the "natural examples" of complete metric spaces without isolated points (of a certain type, or possibly with some additional structure) in fact have at least continuum cardinality. This is not so surprising, since if Cantor's continuum hypothesis holds, then uncountable is equivalent to at least continuum cardinality.

However, if we do not wish to assume CH -- and, ever since Godel and Cohen proved that (G)CH is independent of ZFC set theory, this seems to be the prevalent attitude -- what can be said about the existence of such spaces of uncountable cardinality less than the continuum?

I asked this question to someone before, and I seem to remember that it is known that one cannot unconditionally improve the conclusion of this application of Baire category to say "continuum cardinality". But could someone say a little bit about how this goes? Preferably in words that are comprehensible to a non-set theorist like myself?

Addendum: Thanks to Sergei Ivanov for a quick and convincing answer: evidently I was making things much more complicated than I needed to. Just to get myself reoriented properly, I would like to try to remember where the set-theoretic subtleties come in. Suppose I ask about the conclusion of BCT itself, rather than this particular corollary: not assuming CH, what can we say about the minimal cardinality of a covering family of nowhere dense subsets in a complete metric space?

Second Addendum: I was even more turned around than I had realized: I was (i) worrying needlessly about uncountable cardinals smaller than the continuum and (ii) not worrying enough about cardinals greater than the continuum! In particular, I was under the misimpression that for any cardinal $\kappa \geq 2^{\aleph_0}$, $\kappa^{\aleph_0} = \kappa$. This led me to incorrectly guess the strong form a classical theorem of F.K. Schmidt. I think I have it right now: if you are interested, see pp. 13-16 of

http://math.uga.edu/~pete/8410Chapter3.pdf

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To disclose more fully where I am coming from, I am particularly interested in applications in topological algebra. For instance, recently I was typing up some lecture notes on "multiply complete fields", i.e., fields which are complete with respect to at least two nontrivial rank one valuations. The classical theorem here is due to F.K. Schmidt: a multiply complete field is algebraically closed. So...conversely? It is clear by Baire category that a countable algebraically closed field cannot be complete with respect to even one nontrivial valuation, and on the other hand, if $\kappa$ is any cardinal greater than or equal to $2^{\aleph_0}$, then an algebraically closed field of cardinality $\kappa$ is multiply complete -- e.g. let $k$ be a function field in $\# \kappa$ indeterminates over the prime subfield; there are plenty of inequivalent valuations on $k(t)$, and for any of them, the completion of the algebraic closure is an algebraically closed field of cardinality $\max(2^{\aleph_0},\kappa) = \kappa$, so all these fields are isomorphic. But what about algebraically closed fields of uncountable cardinality less than that of the continuum?

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The Baire category theorem implies that a nonempty complete metric space without isolated points must be uncountable. In many situations I have encountered, the "natural examples" of complete metric spaces without isolated points (of a certain type, or possibly with some additional structure) in fact have at least continuum cardinality. This is not so surprising, since if Cantor's continuum hypothesis holds, then uncountable is equivalent to at least continuum cardinality.

However, if we do not wish to assume CH -- and, ever since Godel and Cohen proved that (G)CH is independent of ZFC set theory, this seems to be the prevalent attitude -- what can be said about the existence of such spaces of uncountable cardinality less than the continuum?

I asked this question to someone before, and I seem to remember that it is known that one cannot unconditionally improve the conclusion of this application of Baire category to say "continuum cardinality". But could someone say a little bit about how this goes? Preferably in words that are comprehensible to a non-set theorist like myself?

To disclose more fully where I am coming from, I am particularly interested in applications in topological algebra. For instance, recently I was typing up some lecture notes on "multiply complete fields", i.e., fields which are complete with respect to at least two nontrivial rank one valuations. The classical theorem here is due to F.K. Schmidt: a multiply complete field is algebraically closed. So...conversely? It is clear by Baire category that a countable algebraically closed field cannot be complete with respect to even one nontrivial valuation, and on the other hand, if $\kappa$ is any cardinal greater than or equal to $2^{\aleph_0}$, then an algebraically closed field of cardinality $\kappa$ is multiply complete -- e.g. let $k$ be a function field in $\# \kappa$ indeterminates over the prime subfield; there are plenty of inequivalent valuations on $k(t)$, and for any of them, the completion of the algebraic closure is an algebraically closed field of cardinality $\max(2^{\aleph_0},\kappa) = \kappa$, so all these fields are isomorphic. But what about algebraically closed fields of uncountable cardinality less than that of the continuum?

Addendum: Thanks to Sergei Ivanov for a quick and convincing answer: evidently I was making things much more complicated than I needed to. Just to get myself reoriented properly, I would like to try to remember where the set-theoretic subtleties come in. Suppose I ask about the conclusion of BCT itself, rather than this particular corollary: not assuming CH, what can we say about the minimal cardinality of a covering family of nowhere dense subsets in a complete metric space?

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